3

Similar to the cfracs in this post, define the two complementary continued fractions,

$$x=\cfrac{-(m+1)}{km\color{blue}+\cfrac{(-1)(2m+1)} {3km\color{blue}+\cfrac{(m-1)(3m+1)}{5km\color{blue} +\cfrac{(2m-1)(4m+1)}{7km\color{blue}+\cfrac{(3m-1)(5m+1)}{9km\color{blue}+\ddots}}}}}\tag1$$

$$y=\cfrac{-(m+1)}{km\color{red}-\cfrac{(-1)(2m+1)} {3km\color{red}-\cfrac{(m-1)(3m+1)}{5km\color{red}-\cfrac{(2m-1)(4m+1)}{7km\color{red}-\cfrac{(3m-1)(5m+1)}{9km\color{red}-\ddots}}}}}\tag2$$

The first one is the superfamily which contains Nicco's cfracs in another post. Let $i$ be the imaginary unit. For $k>1$ and $m>1$, it can be empirically observed that $x$ obeys,

$$\left(\frac{(x+i)^m-(x-i)^m}{(x+i)^m+(x-i)^m}\right) \color{blue}{\left(\frac{(k+i)^{m+1}+(k-i)^{m+1}}{(k+i)^{m+1}-(k-i)^{m+1}}\right)^{(-1)^m}}=1\tag3$$

while $y$ obeys,

$$\left(\frac{(y+1)^m+(y-1)^m}{(y+1)^m-(y-1)^m}\right) \color{blue}{\left(\frac{(k+1)^{m+1}+(k-1)^{m+1}}{(k+1)^{m+1}-(k-1)^{m+1}}\right)^{(-1)^{m+1}}}=-1\tag4$$

where the colored part is a constant that depends on the choice of $k,m$. Hence, as shown in this post, $x,y$ are radicals and algebraic numbers of degree $m$.

Question: How do we prove that $(3)$ and $(4)$ are indeed true?

P.S. Since,

$$\left(\frac{(z+i)^m+(z-i)^m}{2}\right)^2+i^2\left(\frac{(z+i)^m-(z-i)^m}{2}\right)^2 = (z^2+1)^m$$

then the structure of $(3)$ explains the observations about $a^2+b^2=c^m$ in Nicco's post.

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Tito Piezas III
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  • @ Tito Piezas III :Really impressive and magnificent – Nicco May 09 '16 at 21:42
  • @Nicco: Thanks. But I wouldn't have investigated it without your work. :) – Tito Piezas III May 09 '16 at 21:49
  • @ Tito Piezas III: we should also investigate the general form $$F(n,k)=\cfrac{-(m+1)}{km+\cfrac{n(-1)(2m+1)} {3km+\cfrac{n(m-1)(3m+1)}{5km +\cfrac{n(2m-1)(4m+1)}{7km+\cfrac{n(3m-1)(5m+1)}{9km+\ddots}}}}}$$,so that the cfracs in this post become special cases. It also yields algebraic numbers. – Nicco May 11 '16 at 11:29
  • @ Tito Piezas III :I have asked about the general form,please see here – Nicco May 13 '16 at 17:36

2 Answers2

1

Too long for a comment.

If you let $a=-1$ and $b=2m+1$ of the general continued fraction in this post, it reduces to the first continued fraction in this post (with $k=1$) and is expressible as a quotient of gamma functions,

$$x=-\tan\Big(\frac{\pi(m+1)}{4m}\Big)=\frac{\tan\Big(\frac{\pi}{4m}\Big)+1}{\tan\Big(\frac{\pi}{4m}\Big)-1}=-\frac{(m+1)}{4m}\frac{\Gamma\Big(\frac{3m+1}{4m}\Big)\Gamma\Big(\frac{m-1}{4m}\Big)}{\Gamma\Big(\frac{5m+1}{4m}\Big)\Gamma\Big(\frac{3m-1}{4m}\Big)}$$

like you did on the other post. So the conjecture for general $k$ in this post becomes a generalisation.

Tito Piezas III
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Nicco
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  • That's a very nice observation. Since if,$$\frac{R+1}{R-1}=\frac{-a}{b}$$ implies, $$R = -1+\frac{2a}{a+b}$$ we can relate the two cfracs at $k=1$ as, $$\small x=\cfrac{1}{m+\cfrac{(m-1)(m+1)} {3m+\cfrac{(2m-1)(2m+1)}{5m+\cfrac{(3m-1)(3m+1)}{7m+\ddots}}}}\ \small=-1+\cfrac{2(m+1)}{2m+1+\cfrac{(-1)(2m+1)} {3m+\cfrac{(m-1)(3m+1)}{5m+\cfrac{(2m-1)(4m+1)}{7m+\cfrac{(3m-1)(5m+1)}{9m+ \ddots}}}}}$$ where $x = \tan\Big(\frac{\pi}{4m}\Big)$. – Tito Piezas III May 11 '16 at 02:31
0

Using the general solution from this post,we obtain

$$x=\dfrac{ik\sqrt[m]{\dfrac{k+i}{k-i}}-\sqrt[m]{\dfrac{k+i}{k-i}}-ik-1}{k\sqrt[m]{\dfrac{k+i}{k-i}}+i\sqrt[m]{\dfrac{k+i}{k-i}}+k-i}$$

$$y=\dfrac{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}-k-1}{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}+k+1}$$

For $m\gt2$

Nicco
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